Problem: $g(x) = \begin{cases} x^2-5&, & x\in (-\infty,-7) \\\\ 9x-17 &, & x \in [-7,2]\\\\ (x+1)(x-5) &, & x \in (2,\infty)\end{cases}$ $g(7)=$
The strategy First, we should find the appropriate assignment rule out of the three, by checking which case applies for $x={7}$. [I don't understand the notation for the cases.] Finding the appropriate assignment rule Since ${7}\in(2,\infty)$, we should use the third assignment rule $(x+1)(x-5)$. The answer $g({7})=({7}+1)({7}-5)=16$ In conclusion, $g(7)=16$.